5) \( \left\{\begin{array}{l}2 x+3 y \leq 4 \\ x+1 \geq y\end{array}\right. \)

To solve the system of inequalities \[ \left\{ \begin{array}{l} 2x + 3y \leq 4 \\ x + 1 \geq y \end{array} \right. \] we will analyze each inequality separately and then find the region that satisfies both. ### Step 1: Graph the first inequality \(2x + 3y \leq 4\) 1. **Convert to equality**: Start with the equation \(2x + 3y = 4\). 2. **Find intercepts**: - **x-intercept**: Set \(y = 0\): \[ 2x + 3(0) = 4 \implies 2x = 4 \implies x = 2 \quad \text{(point (2, 0))} \] - **y-intercept**: Set \(x = 0\): \[ 2(0) + 3y = 4 \implies 3y = 4 \implies y = \frac{4}{3} \quad \text{(point (0, 4/3))} \] 3. **Draw the line**: Plot the points (2, 0) and (0, 4/3) and draw a line through them. 4. **Shading**: Since the inequality is \( \leq \), shade below the line. ### Step 2: Graph the second inequality \(x + 1 \geq y\) 1. **Convert to equality**: Start with the equation \(y = x + 1\). 2. **Find intercepts**: - **x-intercept**: Set \(y = 0\): \[ 0 = x + 1 \implies x = -1 \quad \text{(point (-1, 0))} \] - **y-intercept**: Set \(x = 0\): \[ y = 0 + 1 = 1 \quad \text{(point (0, 1))} \] 3. **Draw the line**: Plot the points (-1, 0) and (0, 1) and draw a line through them. 4. **Shading**: Since the inequality is \( \geq \), shade above the line. ### Step 3: Find the feasible region The solution to the system of inequalities is the region where the shaded areas from both inequalities overlap. ### Step 4: Identify the vertices of the feasible region To find the vertices of the feasible region, we need to find the intersection of the lines \(2x + 3y = 4\) and \(y = x + 1\). 1. Substitute \(y = x + 1\) into \(2x + 3y = 4\): \[ 2x + 3(x + 1) = 4 \implies 2x + 3x + 3 = 4 \implies 5x + 3 = 4 \implies 5x = 1 \implies x = \frac{1}{5} \] 2. Substitute \(x = \frac{1}{5}\) back into \(y = x + 1\): \[ y = \frac{1}{5} + 1 = \frac{1}{5} + \frac{5}{5} = \frac{6}{5} \] Thus, the intersection point is \(\left(\frac{1}{5}, \frac{6}{5}\right)\). ### Step 5: Identify the vertices of the feasible region The vertices of the feasible region are: - The intersection point \(\left(\frac{1}{5}, \frac{6}{5}\right)\) - The x-intercept of the first inequality (2, 0) - The y-intercept of the first inequality (0, \(\frac{4}{3}\)) - The x-intercept of the second inequality (-1, 0) - The y-intercept of the second inequality (0, 1) ### Conclusion The feasible region is bounded by the points: - \(\left(\frac{1}{5}, \frac{6}{5}\right)\) - (2, 0) - (0, \(\frac{4}{3}\)) - (-1, 0) - (0, 1) You can plot these points to visualize the feasible region. The solution set consists of all points \((x, y)\) that lie within this region.